Two 10 cm long, straight wires, each carrying a current of 5A are kept parallel to each other. If each wire experienced a force of 10⁻⁵ N, then separation between the wires is ____________ cm.

Solution

Physics sir · 4 steps

Use the force on a length L of wire, F = (μ₀ I₁ I₂ L)/(2π d), and solve it for d.

1. Force per length. Each wire sits in the magnetic field of the other. The force on a length L of one wire is set by both currents and the gap d between them.

   F = (μ₀ I₁ I₂ L)/(2π d) ⇒ d = μ₀/2π·(I₁ I₂ L)/F

2. Put in numbers. Use SI units: μ₀/2π = 2 × 10⁻⁷ T m A⁻¹, I₁ I₂ = 5 × 5 = 25 A², L = 0.1 m and F = 10⁻⁵ N.
3. Solve for d. d = (2 × 10⁻⁷ × 25 × 0.1)/10⁻⁵ = (5 × 10⁻⁷)/10⁻⁵ = 0.05 m.
4. So. d = 0.05 m, which is 5 cm.

The common slip: 0.05. Stopping at the SI result gives 0.05, which is the gap in metres. The question asks for centimetres, so multiply by 100.

The idea to keep. For parallel wires, remember μ₀/2π = 2 × 10⁻⁷ in SI units. Put every length in metres, then convert to the unit asked for only at the end.

Answer from the NTA final answer key. Solution by PrepWiser.